Question

The two points on the line 2x + 3y + 4 = 0 which are at distance 2 unit from the line 3x + 4y - 6 = 0 are

• A. (-8, -8) and (-16, -16)
• B. (-44, 64) and (-5, 2)
• C. (-5.5, 1) and (-5,2)
• D. (64, -44) and (4, -4)

Answer 1

The correct option is D (64, -44) and (4, -4)

R.E.F image

Perpendicular distance of point $(x_1,y_1)$

from a line $ax+by+c=0$ is

given by

$\pm d=\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}$

Perpendicular distance $\left(d\right)=2$

Case 1 : for $d=\pm 2$

equation of line : $3x+4y-6=0$

$a=3$ $b=4$ $c=-6$

$\therefore =\frac{3x_1+4y_1-6}{\sqrt{3^2+4^2}}$

$\therefore 10=3x_1+4y_1=6$

$\therefore 3x_1+4y_1-16...\left(1\right)$

Since the point lies on $2ax+3y+4=0$

$2x_1+3y_1=-4...\left(2\right)$

$2x_1=-4-3y_1$

$x_1=-2\frac{-3}{2}{y}_1...\left(2\right)$

Substituting equation (2) in equation (1)

$\therefore 3(-2-\frac{3}{2}{y}_1)+4y_1=16$

$\therefore -6-\frac{9}{2}{y}_1+4y_1=16$

$y_1=-44$

Substituting $y_1=-44$ in (2)

$\therefore x_1=-2-\frac{3}{2}\times (-44)$

$\therefore x_1=64$

$\therefore$ first point is of $(64,-44)$

Case 2 for $d=-2$

$\therefore -2=\frac{3x_1+4y_1-6}{\sqrt{32+42}}$

$\therefore -10=3x_1+4y_1-\mathrm{6...}\left(3\right)$

Substituting equation (2) in equation (3)

$\therefore 3(-2-\frac{3}{2}{y}_1)+4y_1=-4$

$\therefore -6\frac{-9}{2}{y}_1+4y_1=-4$

$y_1=-4$

Substituting the above in equation (2)

$\therefore x_1=-2-\frac{3}{2}\times (-4)$

$\therefore x_1=4$

$\therefore 2^{nd}$ point is $(4,-4)$